Let $$\frac{x^2}{f(a^2+7a+3)} + \frac{y^2}{f(3a+15)} = 1$$ represent an ellipse with major axis along $$y$$-axis, where $$f$$ is a strictly decreasing positive function on $$\mathbf{R}$$. If the set of all possible values of $$a$$ is $$\mathbf{R} - [\alpha, \beta]$$, then $$\alpha^2 + \beta^2$$ is equal to :

This problem is a clever mix of coordinate geometry and functional properties. Here is the step-by-step breakdown.

1. Identify the Ellipse Condition

For the equation $$\frac{x^2}{A} + \frac{y^2}{B} = 1$$ to be an ellipse with its major axis along the $$y$$-axis, the denominator of the $$y^2$$ term must be greater than the denominator of the $$x^2$$ term ($$B > A$$).

From the image, we have:

• $$A = f(a^2 + 7a + 3)$$
• $$B = f(3a + 15)$$
• In a strictly decreasing function, if $$f(x_1) > f(x_2)$$, then $$x_1 < x_2$$.
• $$\alpha = -6$$
• $$\beta = 2$$

So, the condition is:

$$f(3a + 15) > f(a^2 + 7a + 3)$$

2. Apply the Function Property

The problem states that $$f$$ is a strictly decreasing function.

Applying this property to our inequality:

$$3a + 15 < a^2 + 7a + 3$$

3. Solve the Quadratic Inequality

Rearrange the terms to one side:

$$0 < a^2 + 7a - 3a + 3 - 15$$

$$a^2 + 4a - 12 > 0$$

Factor the quadratic:

$$(a + 6)(a - 2) > 0$$

The roots are $$a = -6$$ and $$a = 2$$. For the expression to be greater than zero, $$a$$ must lie outside the roots:

$$a \in (-\infty, -6) \cup (2, \infty)$$

4. Determine $$\alpha$$ and $$\beta$$

The problem defines the set of values as $$\mathbb{R} - [\alpha, \beta]$$.

Our result $$(-\infty, -6) \cup (2, \infty)$$ is equivalent to:

$$\mathbb{R} - [-6, 2]$$

By comparison:

5. Final Calculation

The question asks for the value of $$\alpha^2 + \beta^2$$:

$$\alpha^2 + \beta^2 = (-6)^2 + (2)^2$$

$$36 + 4 = \mathbf{40}$$

Correct Option: B (40)